Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 2}{x + 1} = \dfrac{4x + 3}{x + 1}$
Answer: Multiply both sides by $x + 1$ $ \dfrac{x^2 - 2}{x + 1} (x + 1) = \dfrac{4x + 3}{x + 1} (x + 1)$ $ x^2 - 2 = 4x + 3$ Subtract $4x + 3$ from both sides: $ x^2 - 2 - (4x + 3) = 4x + 3 - (4x + 3)$ $ x^2 - 2 - 4x - 3 = 0$ $ x^2 - 5 - 4x = 0$ Factor the expression: $ (x + 1)(x - 5) = 0$ Therefore $x = -1$ or $x = 5$ However, the original expression is undefined when $x = -1$. Therefore, the only solution is $x = 5$.